Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 43

$${f_{xx}}\left( {x,y} \right) = - \frac{{4y + 2{x^2}{y^2}}}{{{{\left( {2 - {x^2}y} \right)}^2}}}{\text{ and }}{f_{xy}}\left( {x,y} \right) = - \frac{{4x}}{{{{\left( {2 - {x^2}y} \right)}^2}}}$$

$$\eqalign{ & f\left( {x,y} \right) = \ln \left| {2 - {x^2}y} \right| \cr & {\text{find }}{f_x}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\ln \left| {2 - {x^2}y} \right|} \right] \cr & {\text{use the rule }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr & {f_x}\left( {x,y} \right) = \frac{1}{{2 - {x^2}y}}\frac{\partial }{{\partial x}}\left[ {2 - {x^2}y} \right] \cr & {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr & {f_x}\left( {x,y} \right) = \frac{1}{{2 - {x^2}y}}\left( { - 2xy} \right) \cr & {f_x}\left( {x,y} \right) = - \frac{{2xy}}{{2 - {x^2}y}} \cr & \cr & {\text{find }}{f_{xx}}\left( {x,y} \right) \cr & {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - \frac{{2xy}}{{2 - {x^2}y}}} \right] \cr & {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then by using the quotient rule}} \cr & {f_{xx}}\left( {x,y} \right) = - \frac{{\left( {2 - {x^2}y} \right)\frac{\partial }{{\partial x}}\left[ {2xy} \right] - 2xy\frac{\partial }{{\partial x}}\left[ {2 - {x^2}y} \right]}}{{{{\left( {2 - {x^2}y} \right)}^2}}} \cr & {\text{solving derivatives}} \cr & {f_{xx}}\left( {x,y} \right) = - \frac{{\left( {2 - {x^2}y} \right)\left( {2y} \right) - 2xy\left( { - 2xy} \right)}}{{{{\left( {2 - {x^2}y} \right)}^2}}} \cr & {\text{multiply}} \cr & {f_{xx}}\left( {x,y} \right) = - \frac{{4y - 2{x^2}{y^2} + 4{x^2}{y^2}}}{{{{\left( {2 - {x^2}y} \right)}^2}}} \cr & {f_{xx}}\left( {x,y} \right) = - \frac{{4y + 2{x^2}{y^2}}}{{{{\left( {2 - {x^2}y} \right)}^2}}} \cr & \cr & {\text{find }}{f_{xy}}\left( {x,y} \right) \cr & {\text{differentiate }}{f_x}\left( {x,y} \right){\text{ with respect to }}y \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - \frac{{2xy}}{{2 - {x^2}y}}} \right] \cr & {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then by using the quotient rule}} \cr & {f_{xy}}\left( {x,y} \right) = - \frac{{\left( {2 - {x^2}y} \right)\frac{\partial }{{\partial y}}\left[ {2xy} \right] - 2xy\frac{\partial }{{\partial y}}\left[ {2 - {x^2}y} \right]}}{{{{\left( {2 - {x^2}y} \right)}^2}}} \cr & {\text{solving derivatives}} \cr & {f_{xy}}\left( {x,y} \right) = - \frac{{\left( {2 - {x^2}y} \right)\left( {2x} \right) - 2xy\left( { - {x^2}} \right)}}{{{{\left( {2 - {x^2}y} \right)}^2}}} \cr & {\text{multiply}} \cr & {f_{xy}}\left( {x,y} \right) = - \frac{{4x - 2{x^3}y + 2{x^3}y}}{{{{\left( {2 - {x^2}y} \right)}^2}}} \cr & {f_{xy}}\left( {x,y} \right) = - \frac{{4x}}{{{{\left( {2 - {x^2}y} \right)}^2}}} \cr} $$